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Do there exist non-Borel measurable sets?

I think not on $\mathbb{R}$ because open sets, closed sets, compact sets, and so on are Borel since they generate Borel.

J.smith
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    Do you mean a set which is not Borel-measurable, or a set which is measurable but not Borel? – Wojowu Apr 25 '18 at 17:03

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If you use Axiom of choice, for all$y\in (a,b)$, there exists wellorder$<$ and $$A:=\{(x,y) \in (a,b)^2| x<y\}$$ So assume continnum hypothesis, $A_y:=\{x\in (a,b)|x<y \}$ is countable set. Then, by $$\int \int 1_A dx dy =0$$, $$\int \int 1_A dy dx=1$$ So, by Fubini's theorem, $A$ is not measurable.

B.T.M
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    The set $A$ depends on $y$, so I would suggest denoting that somehow (e.g. writing $A_y$). Either way, why would the second integral be $1$? – Wojowu Apr 25 '18 at 17:05
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    Sorry,I mistake. I have edited. – B.T.M Apr 25 '18 at 17:12
  • This is not correct - the axiom of choice is not enough here. Arbitrary well-order will not have the claimed properties. It has to have order type $\omega_1$, and such an order only exists if we assume the continuum hypothesis. – Wojowu Apr 25 '18 at 17:16